Thermodynamics

Kinetic Theory of Gases

Definition

The kinetic theory of gases explains gas pressure as the momentum that molecules deliver to the walls. Add that momentum up and the result is PV = (1/3) N m v_rms^2, with m the mass of one molecule and v_rms their root-mean-square speed. This equation keeps the mass and speed that PV = nRT averages away.

Blow up a balloon and it holds its shape. Nothing inside is pushing the way a hand pushes: the inside is almost entirely empty space, crossed by molecules travelling at hundreds of metres a second. The rubber stays taut because those molecules keep arriving at it and bouncing off.

That is the whole idea, and it buys more than it looks worth. Count the momentum one bounce delivers, multiply by how many bounces arrive each second, and a pressure falls out of pure mechanics — no thermometer, no gas constant, nothing but mass and speed. Then comes the awkward follow-up question this page also answers: how fast is a gas molecule, really?

What Is the Kinetic Theory of Gases?

The kinetic theory of gases treats a gas as a vast number of tiny molecules in constant random motion, and derives its pressure and temperature from their mass, their speed and how often they strike the walls. Everything your instruments report is a statistical summary of that motion.

The model is deliberately crude. It throws away the shape of the molecules, the forces between them and nearly everything else about the substance — and it still predicts real gases accurately over an enormous range of conditions. Five assumptions do the throwing away.

The Wikipedia article on the kinetic theory of gases states them as follows. Each one buys a specific simplification:

  • The particles are very small. The sum of the molecules’ own volume is negligible against the volume of the container, so the gas can be treated as mostly empty space.
  • There are an enormous number of them. The count is large enough that a statistical treatment is justified — the thermodynamic limit — so averages behave like certainties.
  • All collisions are perfectly elastic. Molecules collide with each other and with the walls without losing kinetic energy, which is why a sealed gas never quietly runs down. Our guide to elastic and inelastic collisions covers what that word means in mechanics.
  • Collisions are strictly binary and uncorrelated. No three-body interactions, and the particles have no memory, so every bounce can be worked out on its own.
  • Between collisions the molecules do not interact. They exert no other forces on one another, so each one travels in a straight line at constant speed until it hits something.

Those five lines are what make the arithmetic below possible, and they are also where the model eventually fails — a later section says exactly where. The pressure, volume and temperature relations that drop out at the far end have their own page in our guide to the gas laws.

Brownian motion under a microscope, the visible evidence behind the kinetic theory of gases
Particles jittering under a microscope: every kick comes from molecules far too small to see.

Where Pressure Comes From: The Kinetic Theory Formula

Gas pressure is the rate at which molecules deliver momentum to a surface, and adding that delivery up over a whole container gives the central equation of the kinetic theory.

PV = (1/3) N m v_rms^2
  • P — pressure on the container walls, in pascals (Pa).
  • V — volume of the container, in cubic metres (m3).
  • N — the number of molecules. A plain count, not an amount in moles.
  • m — the mass of one molecule, in kilograms (kg). You get it from the molar mass M in kg/mol as m = M / NA, with NA = 6.02214076e23 per mole. For nitrogen that is 4.65173e-26 kg.
  • v_rms — the root-mean-square molecular speed, in metres per second (m/s): the square root of the mean square speed.
  • rho — the mass density N m / V, in kg/m3, if you prefer the equivalent form P = (1/3) rho v_rms^2.

Watch a single molecule arrive. Its velocity splits into one component along the wall normal and two parallel to the wall, and the bounce reverses only the first of them: the molecule comes in carrying momentum +m vx towards the wall and leaves carrying -m vx, so the wall receives 2 m vx. The two parallel components finish the collision exactly as they started it.

Now for the one third, which is the single thing most often mistaken for a fudge factor. Spread the mean square speed over the three perpendicular directions — v_rms^2 = vx_rms^2 + vy_rms^2 + vz_rms^2 — and equilibrium leaves no direction favoured, so exactly a third belongs to each. Only the axis facing the wall ever reaches the pressure.

Put a number on that. Nitrogen at the ice point, 273.15 K, has an rms speed of 493.169 m/s; divide by sqrt(3) = 1.73205 and the rms speed along any one axis is 284.731 m/s. The one third is a count of dimensions, and problem 2 below works it through.

Run the equation forwards and it needs no temperature at all. Put 2.50e22 nitrogen molecules in a one-litre box at 500.0 m/s and the pressure is 96911.1 Pa, which problem 1 does step by step — or get it instantly from our kinetic theory of gases calculator, which solves PV = (1/3) N m v_rms^2 for whichever of the four quantities you are short of.

One convention, because it moves every figure on this page. Gas states here are quoted at 100 kPa rather than at one standard atmosphere, which is the reference pressure this site’s gas tools use and the reason a mole of gas fills 22.711 litres in them; our guide to Avogadro’s law sets out that choice. At 100 kPa and 273.15 K, a cubic metre holds 2.65165e25 molecules.

Diagram of where the one third in the kinetic theory pressure equation comes from: a gas molecule strikes a container wall, its velocity split into the components vx along the wall normal and vy parallel to the wall, the bounce reversing vx from plus m vx to minus m vx so the wall receives 2 m vx while vy is unchanged; and beside it three orthogonal axes showing that the mean square speed splits equally over x, y and z, so the mean square speed along the one axis facing the wall is a third of the whole, which is the one third in PV = (1/3) N m v_rms^2
One bounce, and the geometry behind the one third: only the component facing the wall reverses, and it carries a third of the mean square speed.
Kinetic Theory of Gases Lab

From Molecules to PV = nRT

The ideal gas law is what the kinetic theory produces the moment you say what temperature is. One substitution gets you there, and nothing else is needed.

PV = (1/3) N m v_rms^2 and v_rms^2 = 3 k T / m, so PV = (1/3) N m (3 k T / m) = N k T

The mass cancels on sight: the m in the spine meets the m underneath 3 k T and disappears. What survives, PV = N k T, is the familiar equation of state written per molecule rather than per mole, and our guide to the ideal gas law is that equation’s own page — the molar form, the four sub-laws hiding inside it and where it stops working.

Notice what the cancellation costs. The spine knows the mass and the speed of the molecules; PV = N k T has swallowed both into a single temperature and will not give them back. Running the same substitution the other way gives the mean translational kinetic energy per molecule, (3/2) k T, which our absolute zero guide uses to follow a gas down towards 0 K.

Here k is the Boltzmann constant, 1.380649e-23 J/K. Since the 2019 redefinition of the SI base units it is exact by definition rather than measured, which is what NIST’s CODATA entry for the Boltzmann constant means by a standard uncertainty marked “exact”. Multiply it by the Avogadro constant and the molar gas constant follows, 8.31446261815324 J/(mol K), which is why R never has to be typed into any tool in this cluster.

The Speed of a Gas Molecule Is Three Different Numbers

A gas has no single molecular speed. Its molecules are spread across a whole distribution of speeds, and three different averages of that distribution are in everyday use — which is why “the” speed of a nitrogen molecule has three defensible answers.

The spread is the Maxwell-Boltzmann speed distribution, the lopsided hump drawn below. Most molecules sit near the peak; a few crawl along near zero, and a thin tail runs out to several times the typical speed. Nothing in the gas sorts them out, because every collision reshuffles who is fast.

Statistic What it measures Formula Nitrogen at 273.15 K As a % of v_rms
Most probable speed, v_p The peak of the curve: the speed more molecules have than any other sqrt(2RT/M) 402.670 m/s 81.65 %
Mean speed, v_bar The plain arithmetic average of every molecule’s speed sqrt(8RT/(pi M)) 454.365 m/s 92.13 %
Root-mean-square speed, v_rms The square root of the mean of the squares — the average the pressure depends on sqrt(3RT/M) 493.169 m/s 100 %

Those percentages are not a nitrogen fact. Wikipedia’s kinetic theory article puts it plainly: “The most probable (or mode) speed v_p is 81.6% of the root-mean-square speed v_rms, and the mean (arithmetic mean, or average) speed [v_bar] is 92.1% of the rms speed (isotropic distribution of speeds).” Change the gas or the temperature and the ratio 1 : 1.12838 : 1.22474 does not move at all.

Why does the pressure need the awkward one? Because the momentum sum produces a mean of squares, not a mean of speeds, and those are different quantities. Squaring before averaging gives the fast molecules extra weight, which is exactly how v_rms ends up 8.54 per cent above the plain mean speed.

Graph of the Maxwell-Boltzmann speed distribution for nitrogen at 273.15 kelvin, molecular speed in metres per second on the horizontal axis from 0 to 1200 and the relative number of molecules per unit speed on the vertical axis with no numeric scale; three vertical markers label the most probable speed 402.670, the mean speed 454.365 and the root-mean-square speed 493.169 metres per second, and the shaded tail to the right of the rms marker is annotated as 39.16 per cent of the molecules
Nitrogen at 273.15 K. The three averages sit close together but never coincide, and the shaded tail beyond the rms speed holds 39.16 per cent of the molecules.

Read the shaded area and 39.16 per cent of the molecules are moving faster than the rms speed. That figure is the same for helium at 100 K and for carbon dioxide at 1000 K, because the shape of the curve depends only on the ratio v / v_p, and v_rms is always sqrt(3/2) times v_p. Push the threshold out to twice the rms speed and 0.738 per cent is left — fewer than one molecule in a hundred.

Why Helium Is Fast and Carbon Dioxide Is Slow

At the same temperature a light molecule moves faster than a heavy one by exactly the square root of the mass ratio, because temperature fixes the kinetic energy per molecule and not the speed. Quadruple the molar mass and you halve the speed.

Every speed on this page carries M under a square root in a denominator, so the ladder at the ice point runs over a factor of 3.3 from the lightest gas to the heaviest:

Gas Molar mass (g/mol) v_rms at 273.15 K (m/s) Speed relative to nitrogen
Helium (He) 4.002602 1304.69 2.6455
Nitrogen (N2) 28.0134 493.169 1.0000
Oxygen (O2) 31.9988 461.436 0.9357
Argon (Ar) 39.948 412.982 0.8374
Carbon dioxide (CO2) 44.0095 393.464 0.7978

The factor between any two of them survives the temperature entirely. Helium’s molecules are sqrt(44.0095 / 4.002602) = 3.31590 times faster than carbon dioxide’s at the ice point, at 1000 K, and at every other temperature the two gases share — the T in the numerator cancels before you reach the square root. Problem 6 takes that apart.

Air is the case worth having. Treated as one average molecule of 28.9645 g/mol it comes out at 485.004 m/s at the ice point, between nitrogen’s figure and oxygen’s, exactly as a mixture of the two should. The mass density the second route needs, 1.27535 kg/m3, is what our gas density calculator returns from rho = P M / (R T), and problem 5 runs the inversion both ways.

One consequence reaches beyond the textbook: sound travels through a gas at a speed comparable to its molecules’, because the same molecules carry it. Our guide to the speed of sound takes that from there.

What the Model Gets You, and What It Does Not

The kinetic theory delivers the pressure, the equation of state and the whole distribution of molecular speeds out of five assumptions, and it loses accuracy in precisely the places where those assumptions stop holding. Two of the five fail first.

Assumption one goes first under compression. Squeeze a gas hard enough and the molecules’ own volume is no longer negligible beside the container’s, so the room they actually have to move in is less than V, and the real pressure runs above what this equation predicts.

Assumption five goes next as the gas cools. Real molecules attract one another weakly at a distance, which takes a little momentum out of each wall strike and pulls the pressure the other way. Where those two corrections become serious is laid out in the ideal gas law guide, which is also where the gas-specific fixes live.

One quantity from this model is deliberately absent here. The mean free path — the average distance a molecule covers between collisions — follows from the number density and a collision cross-section, in the shape 1 / (n sigma sqrt(2)); but a value for sigma has to be measured, and no measured cross-section was sourced for this page. Verify one before use rather than taking a number from here, because there is not one to take.

What you do get, from five lines of assumption, is worth the trade: a pressure out of pure mechanics, an equation of state as a consequence rather than a measurement, and a speed distribution precise enough to name the fraction of a gas above any threshold you care to pick.

Common Misconceptions About the Kinetic Theory of Gases

Misconception 1: gas molecules all move at the same speed

They do not, and the spread is wide. The curve above is the honest picture of one box of nitrogen at one temperature: molecules at a hundred metres a second and molecules at a thousand, simultaneously. Collisions keep reshuffling which molecule is which, so the distribution holds its shape while no individual molecule holds its speed.

Misconception 2: pressure is the gas pushing on the wall

Pressure is momentum delivered per second per unit area, not a steady shove. Between strikes the wall feels nothing at all; the sensation of a constant push comes from the sheer number of arrivals. That distinction is what puts the one third there, since only the velocity component facing the wall ever reverses.

Misconception 3: heavier molecules hit harder, so a heavy gas sits at a higher pressure

At the same temperature and the same number of molecules per cubic metre, every gas sits at exactly the same pressure. Heavier molecules do deliver more momentum per strike, but they are slower by precisely the compensating factor, and the product m v_rms^2 comes out identical. Helium at 1304.69 m/s and carbon dioxide at 393.464 m/s read 100.000 kPa in the same box.

Misconception 4: the rms speed is the average speed

It is 8.54 per cent higher than the average. Squaring before averaging weights the fast molecules more heavily than a plain arithmetic mean does, so v_rms sits above v_bar for every gas at every temperature — 493.169 against 454.365 m/s for nitrogen at the ice point. Our absolute zero guide makes the same point about that 8.5 per cent gap.

Worked Problems

Problem 1
A sealed 1.00 L box holds 2.50e22 nitrogen molecules whose rms speed is 500.0 m/s. Find the pressure inside, and the temperature that speed implies. Nitrogen's molar mass is 28.0134 g/mol.
Show Solution
Solution: Step 1: Use the spine equation in its number-density form, P = (1/3) (N / V) m v_rms^2. The m in it is the mass of ONE molecule, which is the molar mass divided by the Avogadro constant: m = M / NA. Step 2: m = 0.0280134 kg/mol / 6.02214076e23 per mol = 4.651735e-26 kg. Carry that seventh figure: it is 4.65173e-26 kg to six, and the last digit of the pressure depends on which one you keep. Step 3: The number density is N / V = 2.50e22 / 1.00e-3 m3 = 2.50e25 per m3, and the whole mass of gas in the box is N m = 2.50e22 x 4.651735e-26 kg = 0.00116293 kg. Step 4: P = (1/3) x 2.50e25 per m3 x 4.651735e-26 kg x (500.0 m/s)^2 = 96911.1 Pa. Step 5: The temperature follows from v_rms = sqrt(3 k T / m) rearranged: T = m v_rms^2 / (3 k) = 4.651735e-26 x (500.0)^2 / (3 x 1.380649e-23) = 280.770 K. Put that temperature into N k T / V and the same pressure comes back, which checks the algebra rather than adding a second claim. Answer: 96911.1 Pa, which is 96.9111 kPa or 0.9564 atm, at an implied temperature of 280.770 K. That pressure came out of a mass and a speed with no thermometer anywhere in it. And because v_rms is squared, the same gas moving 10 per cent faster sits at 1.210 times this pressure, not 1.10 times.
Problem 2
Nitrogen sits at 273.15 K. Work out its rms speed, then the rms speed along a single axis, and say why that puts a one third in the pressure equation. Molar mass 28.0134 g/mol, R = 8.31446261815324 J/(mol K).
Show Solution
Solution: Step 1: The rms speed comes from v_rms = sqrt(3 R T / M), the molar form of sqrt(3 k T / m). Step 2: v_rms = sqrt(3 x 8.31446261815324 x 273.15 / 0.0280134) = 493.169 m/s. Step 3: The mean square speed splits over three perpendicular axes, v_rms^2 = vx_rms^2 + vy_rms^2 + vz_rms^2, and at equilibrium the three terms are equal, so vx_rms^2 = v_rms^2 / 3 and vx_rms = v_rms / sqrt(3), with sqrt(3) = 1.73205. Step 4: vx_rms = 493.169 m/s / 1.73205 = 284.731 m/s. Answer: 493.169 m/s overall, and 284.731 m/s along any one axis. Only that single axis reverses at the wall, so only a third of the mean square speed is available to the pressure. The one third is a count of dimensions, not a correction factor fitted to data.
Problem 3
Helium and nitrogen are held at the same temperature, 273.15 K. Find the rms speed of each and the factor between them. Molar masses: helium 4.002602 g/mol, nitrogen 28.0134 g/mol.
Show Solution
Solution: Step 1: Both come from v_rms = sqrt(3 R T / M), with R = k NA = 8.31446261815324 J/(mol K). Step 2: For helium, 3 R T / M = 3 x 8.31446261815324 x 273.15 / 0.004002602 = 1702214 m2/s2, so v_rms = sqrt(1702214) = 1304.69 m/s. Step 3: Nitrogen is problem 2’s figure, 493.169 m/s. The factor between them is the inverse square root of the mass ratio: sqrt(28.0134 / 4.002602) = sqrt(6.99880) = 2.6455. Answer: helium 1304.69 m/s, nitrogen 493.169 m/s, a factor of 2.6455. Temperature fixes the kinetic energy per molecule, not the speed, so the lighter molecule has to move faster to carry the same energy.
Problem 4
For nitrogen at 273.15 K, work out the most probable speed, the mean speed and the rms speed, then express the first two as percentages of the third.
Show Solution
Solution: Step 1: The three averages of the same distribution are v_p = sqrt(2 R T / M), v_bar = sqrt(8 R T / (pi M)) and v_rms = sqrt(3 R T / M). Only the numerical factor inside the root changes. Step 2: v_p = sqrt(2 x 8.31446261815324 x 273.15 / 0.0280134) = 402.670 m/s, and v_bar = sqrt(8 x 8.31446261815324 x 273.15 / (pi x 0.0280134)) = 454.365 m/s. The rms speed is problem 2’s 493.169 m/s. Step 3: As fractions of the rms speed, 402.670 / 493.169 = 0.8165 and 454.365 / 493.169 = 0.9213. Answer: v_p = 402.670 m/s, v_bar = 454.365 m/s and v_rms = 493.169 m/s, which makes the first two 81.65 % and 92.13 % of the third. Those two percentages belong to the shape of the distribution rather than to nitrogen: every gas at every temperature returns the same pair.
Problem 5
Dry air sits at 100 kPa and 273.15 K and behaves as one average molecule of molar mass 28.9645 g/mol. Find its mass density, then get its rms speed from the pressure and density alone, and check that against sqrt(3RT/M).
Show Solution
Solution: Step 1: Take the density from rho = P M / (R T), then invert the density form of the spine, P = (1/3) rho v_rms^2, into v_rms = sqrt(3 P / rho). Step 2: rho = 100000 x 0.0289645 / (8.31446261815324 x 273.15) = 1.275354 kg/m3, which is 1.27535 kg/m3 to six figures. Step 3: v_rms = sqrt(3 x 100000 / 1.275354) = 485.004 m/s. Step 4: The independent route needs no density at all: sqrt(3 R T / M) = sqrt(3 x 8.31446261815324 x 273.15 / 0.0289645) = 485.004 m/s. Answer: rho = 1.27535 kg/m3, and v_rms = 485.004 m/s by both routes. Carry the unrounded density into the root the way step 3 does; starting from the six-figure 1.27535 kg/m3 instead returns 485.005 m/s. The first route needs a pressure and a density, the pair you can measure without a thermometer; the second needs a temperature and a molar mass and never mentions the pressure. They meet at the same figure, which is what makes the agreement worth printing.
Problem 6
By what factor is a helium molecule faster than a carbon dioxide molecule at the same temperature, and to what temperature would carbon dioxide have to be heated to match helium at 273.15 K? Molar masses: helium 4.002602 g/mol, carbon dioxide 44.0095 g/mol.
Show Solution
Solution: Step 1: Divide sqrt(3 R T / M) for one gas by the same expression for the other at the same T. The 3, the R and the T all cancel, leaving v_rms(He) / v_rms(CO2) = sqrt(M_CO2 / M_He). Step 2: M_CO2 / M_He = 44.0095 / 4.002602 = 10.9952, so the factor is sqrt(10.9952) = 3.31590. No temperature survives in that expression, so it holds at every temperature the two gases share. Step 3: Matching helium’s speed means matching T / M. Carbon dioxide therefore needs T = 273.15 K x 44.0095 / 4.002602 = 3003.35 K. Answer: 3.31590 times faster at any shared temperature, and carbon dioxide would have to reach 3003.35 K. Equal molecular speeds mean equal T / M, never equal T. That is the whole reason helium escapes a balloon while carbon dioxide loiters.
Problem 7
Open the simulator above, press the preset Ice point - N2 at 273.15 K, and leave the threshold slider at 1.00 x v_rms. Read the cards, then check them against each other.
Show Solution
Solution: Step 1: The preset pins nitrogen at 273.15 K and the anchor number density, 2.65165e+25 per m3. The three speed cards read 402.670 m/s most probable, 454.365 m/s mean and 493.169 m/s rms, and the pressure card reads 100.000 kPa. Step 2: The threshold row reads 1.00 x v_rms = 493.169 m/s and 39.16 %, and the ratio row reads 1 : 1.12838 : 1.22474. Step 3: Check the cards against one another: 493.169 / 454.365 = 1.08540, so the rms speed stands 8.54 per cent above the mean speed. Now press the He and CO2 buttons in turn. Every speed moves, the pressure does not, and the 39.16 % never moves at all. Answer: v_p 402.670 m/s, v_bar 454.365 m/s, v_rms 493.169 m/s, P = 100.000 kPa, and 39.16 % of the molecules faster than the rms speed. Those are the same three speeds problem 4 computed by hand, which is exactly the point of having both.

Frequently Asked Questions

What is the kinetic theory of gases in simple terms?
It is the idea that a gas is nothing but an enormous number of tiny molecules flying about at random, and that everything you can measure about the gas is an average over their motion. Pressure is the momentum they deliver to the walls, and temperature is a measure of their kinetic energy. Add the momentum up and the gas laws follow as a consequence.
What are the assumptions of the kinetic theory of gases?
There are five. The molecules are small enough that their own volume is negligible against the container; there are enough of them for statistics to be reliable; all collisions are perfectly elastic; collisions are strictly binary and uncorrelated, with no memory; and between collisions the molecules exert no forces on one another. Those five are what make the pressure sum possible at all.
What is the formula for the kinetic theory of gases?
PV = (1/3) N m v_rms^2, where P is the pressure, V the volume, N the number of molecules, m the mass of one molecule and v_rms the root-mean-square molecular speed. Written with the mass density rho = N m / V it becomes P = (1/3) rho v_rms^2. Substituting v_rms^2 = 3 k T / m turns the same equation into PV = N k T.
Why is there a 1/3 in PV = (1/3) N m v_rms^2?
Because a molecule’s mean square speed is shared equally between three perpendicular directions, and only the one facing the wall delivers momentum to it. The pressure therefore sees v_rms^2 / 3 rather than the whole of v_rms^2. That one third is a count of dimensions rather than an empirical correction, which is why it is exactly a third and not a fitted number.
What is the difference between the mean speed and the rms speed?
The mean speed is the plain arithmetic average of every molecule’s speed; the rms speed is the square root of the average of their squares. Squaring first gives the fast molecules extra weight, so the rms speed always comes out larger, by 8.54 per cent, for every gas at every temperature. The pressure equation needs the rms one because the momentum sum produces a mean of squares.
How fast do air molecules actually move?
At the ice point, 273.15 K, nitrogen’s rms speed is 493.169 m/s and oxygen’s is 461.436 m/s, so air molecules sit near 500 metres per second. Treating dry air as one average molecule of 28.9645 g/mol gives 485.004 m/s. All three figures are quoted at that temperature; raise it, and every one of them climbs as the square root of T.
Does a heavier gas exert more pressure?
No. At the same temperature, with the same number of molecules in the same volume, every gas sits at exactly the same pressure. Heavier molecules carry more momentum into each strike, but they travel more slowly by precisely the compensating factor, so the product m v_rms^2 is identical. Helium at 1304.69 m/s and carbon dioxide at 393.464 m/s both read 100.000 kPa.
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